Advertisements
Advertisements
Question
The average energy in one time period in simple harmonic motion is
Options
\[\frac{1}{2}m \omega^2 A^2\]
\[\frac{1}{4}m \omega^2 A^2\]
m ω2A2
zero
Advertisements
Solution
\[\frac{1}{2}m \omega^2 A^2\]
It is the total energy in simple harmonic motion in one time period.
APPEARS IN
RELATED QUESTIONS
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
(a) a = 0.7x
(b) a = –200x2
(c) a = –10x
(d) a = 100x3
A particle executes S.H.M. with a period of 10 seconds. Find the time in which its potential energy will be half of its total energy.
State the differential equation of linear simple harmonic motion.
Can simple harmonic motion take place in a non-inertial frame? If yes, should the ratio of the force applied with the displacement be constant?
The time period of a particle in simple harmonic motion is equal to the time between consecutive appearances of the particle at a particular point in its motion. This point is
The displacement of a particle is given by \[\overrightarrow{r} = A\left( \overrightarrow{i} \cos\omega t + \overrightarrow{j} \sin\omega t \right) .\] The motion of the particle is
The motion of a torsional pendulum is
(a) periodic
(b) oscillatory
(c) simple harmonic
(d) angular simple harmonic
Suppose a tunnel is dug along a diameter of the earth. A particle is dropped from a point, a distance h directly above the tunnel. The motion of the particle as seen from the earth is
(a) simple harmonic
(b) parabolic
(c) on a straight line
(d) periodic
A particle moves on the X-axis according to the equation x = x0 sin2 ωt. The motion is simple harmonic
The angle made by the string of a simple pendulum with the vertical depends on time as \[\theta = \frac{\pi}{90} \sin \left[ \left( \pi s^{- 1} \right)t \right]\] .Find the length of the pendulum if g = π2 m2.
A small block oscillates back and forth on a smooth concave surface of radius R in Figure. Find the time period of small oscillation.

A particle executing SHM crosses points A and B with the same velocity. Having taken 3 s in passing from A to B, it returns to B after another 3 s. The time period is ____________.
The length of a second’s pendulum on the surface of the Earth is 0.9 m. The length of the same pendulum on the surface of planet X such that the acceleration of the planet X is n times greater than the Earth is
Define the time period of simple harmonic motion.
A simple harmonic motion is given by, x = 2.4 sin ( 4πt). If distances are expressed in cm and time in seconds, the amplitude and frequency of S.H.M. are respectively,
Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is ______.
- simple harmonic motion.
- non-periodic motion.
- periodic motion.
- periodic but not S.H.M.
Displacement vs. time curve for a particle executing S.H.M. is shown in figure. Choose the correct statements.

- Phase of the oscillator is same at t = 0 s and t = 2s.
- Phase of the oscillator is same at t = 2 s and t = 6s.
- Phase of the oscillator is same at t = 1 s and t = 7s.
- Phase of the oscillator is same at t = 1 s and t = 5s.
Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock - 2 is placed on a space station located at a height h above the earth's surface. Clock - 1 and Clock - 2 operate at time periods 4 s and 6 s respectively. Then the value of h is ______.
(consider the radius of earth RE = 6400 km and g on earth 10 m/s2)
